After the closing of the mill, the town of Sawyerville experienced a decline in its population. The relationship between the elapsed time, $t$, in years, since the closing of the mill, and the town's population, $P(t)$, is modeled by the following function. P ( t ) = 12,000 ⋅ 2 − t 15 P(t)=12{,}000\cdot 2\^{{-\frac{t}{15}}} In how many years will Sawyerville's population be 9000? Round your answer, if necessary, to the nearest hundredth.
Answer: Thinking about the problem We want to know how many years, $t$, it will take for the population of Sawyerville, $P(t)$, to be $9000$. So we need to find the value of $t$ for which $P(t)=9000$. Substituting $9000$ in for $P(t)$ in the function gives us the following equation. 9000 = 12,000 ⋅ 2 − t 15 9000=12{,}000\cdot 2\^{{-\frac{t}{15}}} Solving the equation We can solve the equation as shown below. 12,000 ⋅ 2 − t 15 2 − t 15 − t 15 t = 9000 = 0.75 = log 2 ( 0.75 ) = − 15 log 2 ( 0.75 ) \begin{aligned}12{,}000\cdot 2\^{{-\frac{t}{15}} }&=9000\\\\ 2\^{{-\frac{t}{15}} }&=0.75\\\\ -\dfrac{t}{15}&=\log_2(0.75)\\\\ t&=-15\log_2(0.75) \end{aligned} Changing the base to approximate the solution Since most calculators only calculate logarithms in base $10$ and base $e$, let's change the base. [What is the change of base rule?] $\begin{aligned}t&=-15\log_2(0.75)\\\\ &=-15\cdot \dfrac{\log(0.75)}{\log(2)}\\\\ &\approx 6.23 \end{aligned}$ In about $6.23$ years, the population of Sawyerville will be $9000$.